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vfiekz [6]
3 years ago
12

A rectangular current loop is located near a long straight wire that carries a current of 12A. The current in the loop is 5A. De

termine the magnitude of the net magnetic force that acts on the loop...
Physics
1 answer:
masya89 [10]3 years ago
8 0

Explanation:

The net magnetic force acting on the loop is the superposition of the forces on each side of the loop. The forces on the two sides perpendicular to the long straight wire will cancel (right hand rule). The magnitude of the magnetic force between two wires with currents I_1 = 12 A and I_2 = 5 A of length L = 0.5 m ( assuming not given in question).

F = \frac{\mu_o I_1 I_2 L}{2 \pi d}

so the forces on the near and far wires are

F_n = 2.7\times 10^{-4} N

F_f = 1.2\times 10^{-4} N

The force F_n is an attractive one, while F_f is a repulsive one (right hand rule). The magnitude of the force is therefore

F = F_n - F_f

F=1.5\times10^{-4}

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A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150
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The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

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b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

\rho \mathrm{Vg}=2 \mathrm{mg} \\

V=\frac{2 m}{\rho} \\

V=\frac{2 \times 150}{1000} \\

\mathrm{~V}=0.3 \mathrm{~m}^3

Area of pool $=\frac{\pi}{4} d^2$

&=\frac{\pi}{4} \times 6^2 \\

&=28.27 \mathrm{~m}^2

Height of the water rise

h &=\frac{V}{A} \\

h &=\frac{0.3}{28.27} \\

& \mathrm{~h}=0.0106 \mathrm{~m}

  • Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

brainly.com/question/17200230

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4 0
2 years ago
Solve the question that follows using the equation for the conversion of Celsius to Fahrenheit. F=95(C)+32 On February 9, 1934,
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So, the temperature in Buffalo, NY was -20.2°F and the temperature in Anchorage, AL was 19°F.

Hence, it was colder in Buffalo, NY than in Anchorage, AL.

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Explanation:

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